Punnett Square Calculator
Draw the Punnett square for any cross of one, two or three genes, or an X-linked cross. Get every genotype and phenotype as an exact fraction and a percentage, the ratios such as 9:3:3:1, with complete, incomplete or codominance for each gene.
How to Use
- Choose Autosomal for ordinary genes, or X-linked for a gene on the X chromosome.
- Type each parent’s genotype, two letters per gene: Aa, AaBb or AaBbCc. Capital letters are dominant alleles. For X-linked, write the mother as XAXa (or XᴬXᵃ) and the father as XAY.
- Set the dominance of each gene: complete (Aa looks like AA), incomplete (Aa is a blend) or codominant (both alleles show).
- Optionally enter a genotype, such as aabb, to get its exact probability.
- Read the ratios in the readouts and the grid, with every genotype and phenotype as a fraction and percentage below it; Show Work lists the gametes and the product rule.
Worked Example
Monohybrid cross. Two heterozygous pea plants, Aa × Aa. Each makes A and a gametes in equal numbers, so the four boxes are AA, Aa, Aa and aa: genotypes 1:2:1, and with A dominant, phenotypes 3:1. Each offspring has a 1/4 = 25% chance of being aa.
Dihybrid cross. AaBb × AaBb: each parent makes AB, Ab, aB and ab, giving a 16-box grid. The phenotypes are 9 A_B_ : 3 A_bb : 3 aaB_ : 1 aabb, so a double-recessive offspring has a 1/16 = 6.25% chance, which is the product 1/4 × 1/4.
The common mistake: adding probabilities that should be multiplied. P(aa) = 1/4 and P(bb) = 1/4, but P(aabb) is not 1/4 + 1/4 = 1/2: both things must happen together, so it is 1/4 × 1/4 = 1/16. Adding is for “either one or the other” of outcomes that cannot happen together, such as P(AA or Aa) = 1/4 + 1/2 = 3/4. The other classic slip is reading 9:3:3:1 as the genotype ratio; it is the phenotype ratio, and there are 9 genotypes in the ratio 1:2:1:2:4:2:1:2:1.
Show Work
Formulas
From Mendel’s Peas to Punnett’s Grid
Gregor Mendel, an Augustinian friar in Brno, published his pea-breeding experiments in 1866. Crossing hybrids for seed shape, he counted 5,474 round and 1,850 wrinkled seeds, a ratio of 2.96 to 1, close to the 3:1 his theory of paired factors predicted. His work went almost unnoticed until 1900, when Hugo de Vries, Carl Correns and Erich von Tschermak each reached the same conclusions.
The grid itself is named after Reginald Punnett, a Cambridge geneticist who worked with William Bateson in the early 1900s and introduced the square as a quick way to list every combination of gametes. Correns also described intermediate (incomplete) inheritance of flower colour, and in 1910 Thomas Hunt Morgan’s white-eyed fruit flies showed a trait following the X chromosome, the pattern the X-linked mode reproduces.
About This Tool
This calculator builds the full Punnett square for crosses of up to three genes, so 2 × 2, 4 × 4 and 8 × 8 grids, and for X-linked crosses where sons and daughters inherit differently. Every probability is an exact fraction, worked with whole-number arithmetic rather than rounded decimals, alongside its percentage, and you can set complete dominance, incomplete dominance or codominance separately for each gene. Show Work lists the gametes, every genotype and phenotype count, and the product-rule check.
Everything runs in your browser; nothing you enter is sent anywhere. It assumes the genes assort independently (no linkage).
Related tools: Hardy–Weinberg Calculator, Probability Calculator, and Fraction Calculator.
Frequently Asked Questions
How does a Punnett square work?
List each parent’s possible gametes along the top and side, then fill each box with the combination. For Aa × Aa the gametes are A and a from each parent, giving AA, Aa, Aa and aa: a 1:2:1 genotype ratio and, with complete dominance, a 3:1 phenotype ratio. Each box is one quarter, so the chance of aa is 1/4 = 25%.
Where does the 9:3:3:1 ratio come from?
A dihybrid cross AaBb × AaBb has 4 gametes per parent and a 4 × 4 = 16-box grid. Because the genes assort independently, the phenotypes multiply: A_B_ = 3/4 × 3/4 = 9/16 (56.25%), A_bb and aaB_ = 3/16 each, and aabb = 1/4 × 1/4 = 1/16 (6.25%).
What is the difference between incomplete dominance and codominance?
In both, the heterozygote looks different from either homozygote, so Rr × Rr gives a 1:2:1 phenotype ratio instead of 3:1. In incomplete dominance the heterozygote is a blend, like pink flowers from red and white parents; in codominance both alleles show fully, like the AB blood group, where the A and B antigens are both present.
How do X-linked traits pass from parents to children?
A son gets his only X from his mother, so for a carrier mother (XᴬXᵃ) and an unaffected father (XᴬY), 1/2 of sons have the recessive trait and 1/2 of daughters are carriers. Across all children that is 1/4 affected sons, 1/4 carrier daughters, 1/4 unaffected sons and 1/4 non-carrier daughters.
How big is a trihybrid Punnett square?
AaBbCc × AaBbCc has 8 gametes per parent, so 8 × 8 = 64 boxes and 27 different genotypes. The phenotypes come out 27:9:9:9:3:3:3:1; the fully recessive aabbcc is 1/64 (1.5625%) and the triple heterozygote AaBbCc is 8/64 = 1/8.
How do I use the Punnett Square Calculator?
Just type your numbers. The answer shows up right away — there is no button to press. Change anything and it updates by itself.
Do I need to install or sign up for anything?
Not at all — it runs in the browser with nothing to install and no account. After it loads once, it even works without an internet connection.
Is my information private?
Yes. Everything happens in your browser. Nothing you type is sent to a server or saved anywhere.
Common Use Cases
Biology homework
Check a dihybrid cross: AaBb × AaBb gives 9 A_B_ : 3 A_bb : 3 aaB_ : 1 aabb, with genotypes 1:2:1:2:4:2:1:2:1.
Test crosses
Crossing AaBb with aabb gives four phenotypes at 1/4 each (1:1:1:1), which reveals the hidden alleles of the first parent.
Family risk questions
A carrier mother and an unaffected father have a 1/2 chance that each son shows an X-linked recessive trait.
Plant and animal breeding
With incomplete dominance, Rr × Rr gives 1/4 red, 1/2 blended and 1/4 white offspring, so every class can be told apart.
Probability practice
The chance of aabbcc from two triple heterozygotes is 1/4 × 1/4 × 1/4 = 1/64, about 1.56%.
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